Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $a = \dfrac{n^2 - 2n - 24}{5n^2 + 45n} \times \dfrac{2n + 18}{-5n + 30} $
Solution: First factor the quadratic. $a = \dfrac{(n - 6)(n + 4)}{5n^2 + 45n} \times \dfrac{2n + 18}{-5n + 30} $ Then factor out any other terms. $a = \dfrac{(n - 6)(n + 4)}{5n(n + 9)} \times \dfrac{2(n + 9)}{-5(n - 6)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (n - 6)(n + 4) \times 2(n + 9) } { 5n(n + 9) \times -5(n - 6) } $ $a = \dfrac{ 2(n - 6)(n + 4)(n + 9)}{ -25n(n + 9)(n - 6)} $ Notice that $(n + 9)$ and $(n - 6)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 2\cancel{(n - 6)}(n + 4)(n + 9)}{ -25n(n + 9)\cancel{(n - 6)}} $ We are dividing by $n - 6$ , so $n - 6 \neq 0$ Therefore, $n \neq 6$ $a = \dfrac{ 2\cancel{(n - 6)}(n + 4)\cancel{(n + 9)}}{ -25n\cancel{(n + 9)}\cancel{(n - 6)}} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $a = \dfrac{2(n + 4)}{-25n} $ $a = \dfrac{-2(n + 4)}{25n} ; \space n \neq 6 ; \space n \neq -9 $